Olympiad problems

2016 Denmark MO - Mohr Contest, 3

Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]

2015 Saudi Arabia JBMO TST, 3

Tags: circumcircle , right triangle , right angle , geometry

A right triangle $ABC$ with $\angle C=90^o$ is inscribed in a circle. Suppose that $K$ is the midpoint of the arc $BC$ that does not contain $A$. Let $N$ be the midpoint of the segment $AC$, and $M$ be the intersection point of the ray $KN$ and the circle.The tangents to the circle drawn at $A$ and $C$ meet at $E$. prove that $\angle EMK = 90^o$

2018 NZMOC Camp Selection Problems, 4

Tags: geometry , right angle , equal angles

Let $P$ be a point inside triangle $ABC$ such that $\angle CPA = 90^o$ and $\angle CBP = \angle CAP$. Prove that $\angle P XY = 90^o$, where $X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.

2002 IMO Shortlist, 8

Tags: geometry , circles , right angle

Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$. A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$. Let $M$, $N$, $K$ be three points on the line segments $CD$, $BC$, $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$.

1978 Vietnam National Olympiad, 5

Tags: geometry , maximum , right angle

A river has a right-angle bend. Except at the bend, its banks are parallel lines of distance $a$ apart. At the bend the river forms a square with the river flowing in across one side and out across an adjacent side. What is the longest boat of length $c$ and negligible width which can pass through the bend?

Kharkiv City MO Seniors - geometry, 2014.11.5

Tags: right angle , geometry , collinear , perpendicular

In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

Tags: geometry , equal segments , equal angles , right angle

Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$. (Anton Trygub)

2021 Dutch IMO TST, 2

Tags: right angle , centroid , right triangle , geometry

Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.

2017 Bosnia and Herzegovina Junior BMO TST, 3

Tags: incircle , right angle , incenter , geometry

Let $ABC$ be a triangle such that $\angle ABC = 90 ^{\circ}$. Let $I$ be an incenter of $ABC$ and let $F$, $D$ and $E$ be points where incircle touches sides $AB$, $BC$ and $AC$, respectively. If lines $CI$ and $EF$ intersect at point $M$ and if $DM$ and $AB$ intersect in $N$, prove that $AI=ND$

2020 Tournament Of Towns, 5

Tags: right angle , trapezoid , equal segments , trapezium , geometry

Let $ABCD$ be an inscribed trapezoid. The base $AB$ is $3$ times longer than $CD$. Tangents to the circumscribed circle at the points $A$ and $C$ intersect at the point $K$. Prove that the angle $KDA$ is a right angle. Alexandr Yuran

Croatia MO (HMO) - geometry, 2020.3

Tags: geometry , right angle , perpendicular

Given a triangle $ABC$ such that $AB<AC$ . On sides $AB$ and $BC$, points $P$ and $Q$ are marked respectively such that the lines $AQ$ and $CP$ are perpendicular, and the circle inscribed in the triangle $ABC$ touches the length $PQ$. The line $CP$ intersects the circle circumscribed around the triangle $ABC$ at the points $C$ and $T$. If the lines $CA,PQ$ and $BT$ intersect at one point, prove that the angle $\angle CAB$ is right.

1992 Swedish Mathematical Competition, 5

Tags: right triangle , right angle , circumradius , geometry

A triangle has sides $a, b, c$ with longest side $c$, and circumradius $R$. Show that if $a^2 + b^2 = 2cR$, then the triangle is right-angled.

2018 Dutch IMO TST, 3

Tags: geometry , right angle , equal segments

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satises $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.

2019 Saudi Arabia Pre-TST + Training Tests, 3.2

Tags: geometry , right angle , circles

Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB,AC$ at $F,E$ respectively. Let $P$ a point on this circle. Let $C',B$' be the projections of $P$ upon the sides $AB,AC$ respectively. Let $H$ be the orthocenter of the triangle $AB'C'$. Show that $\angle EHF = 90^o$.

Indonesia MO Shortlist - geometry, g9

Tags: right angle , arc midpoint , circles , geometry

Given two circles $\Gamma_1$ and $\Gamma_2$ which intersect at points $A$ and $B$. A line through $A$ intersects $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$, respectively. Let $M$ be the midpoint of arc $BC$ in $\Gamma_1$ ,which does not contains $A$, and $N$ is the midpoint of the arc $BD$ in $\Gamma_2$, which does not contain $A$. If $K$ is the midpoint of $CD$, prove that $\angle MKN = 90^o.$

2017 Junior Regional Olympiad - FBH, 5

Tags: geometry , midpoint , right angle

Points $K$ and $L$ are on side $AB$ of triangle $ABC$ such that $KL=BC$ and $AK=LB$. Let $M$ be a midpoint of $AC$. Prove that $\angle KML = 90^{\circ}$

2011 Saudi Arabia IMO TST, 2

Tags: geometry , incenter , circumcircle , right angle

Let $ABC$ be a triangle with $AB\ne AC$. Its incircle has center $I$ and touches the side $BC$ at point $D$. Line $AI$ intersects the circumcircle $\omega$ of triangle $ABC$ at $M$ and $DM$ intersects again $\omega$ at $P$. Prove that $\angle API= 90^o$.

2021 Polish Junior MO Second Round, 2

Tags: right angle , square , geometry

Given is the square $ABCD$. Point $E$ lies on the diagonal $AC$, where $AE> EC$. On the side $AB$, a different point from $B$ has been selected for which $EF = DE$. Prove that $\angle DEF = 90^o$.

2019 Grand Duchy of Lithuania, 3

Tags: geometry , right angle , perpendicular , orthocenter , circumcircle

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The perpendicular bisector of segment $CH$ intersects the sides $AC$ and $BC$ in points $X$ and $Y$ , respectively. The lines $XO$ and $YO$ intersect the side $AB$ in points $P$ and $Q$, respectively. Prove that if $XP + Y Q = AB + XY$ then $\angle OHC = 90^o$.

1998 Tuymaada Olympiad, 8

Tags: geometry , 3d geometry , pyramid , right angle , solid geometry

Given the pyramid $ABCD$. Let $O$ be the midpoint of edge $AC$. Given that $DO$ is the height of the pyramid, $AB=BC=2DO$ and the angle $ABC$ is right. Cut this pyramid into $8$ equal and similar to it pyramids.

2020 China Team Selection Test, 2

Tags: geometry , right angle , equal segments , tangent circle

Given an isosceles triangle $\triangle ABC$, $AB=AC$. A line passes through $M$, the midpoint of $BC$, and intersects segment $AB$ and ray $CA$ at $D$ and $E$, respectively. Let $F$ be a point of $ME$ such that $EF=DM$, and $K$ be a point on $MD$. Let $\Gamma_1$ be the circle passes through $B,D,K$ and $\Gamma_2$ be the circle passes through $C,E,K$. $\Gamma_1$ and $\Gamma_2$ intersect again at $L \neq K$. Let $\omega_1$ and $\omega_2$ be the circumcircle of $\triangle LDE$ and $\triangle LKM$. Prove that, if $\omega_1$ and $\omega_2$ are symmetric wrt $L$, then $BF$ is perpendicular to $BC$.

2019 New Zealand MO, 2

Tags: geometry , angle bisector , right angle

Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^o$. Prove that $QP$ is the angle bisector of $\angle BQC$.

2021 Dutch IMO TST, 2

Tags: right angle , centroid , right triangle , geometry

Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.

Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Tags: right angle , angle , geometry

Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Inside $\vartriangle BMC$ was found a point $P$ such that $\angle BMP = 90^o$, $\angle ABC+ \angle APC =180^o$. Prove that $\angle PBM + \angle CBM = \angle PCA$. (Anton Trygub)

2017 Auckland Mathematical Olympiad, 1

Tags: geometry , right angle

A $6$ meter ladder rests against a vertical wall. The midpoint of the ladder is twice as far from the ground as it is from the wall. At what height on the wall does the ladder reach?